The angle between the tangents drawn from the | vedclass

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(C) Let the angle between the tangents be $2	heta = \cos^{-1}\left(\frac{7}{16}ight)$.Then $\cos(2	heta) = \frac{7}{16}$.Using the identity $\cos(

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$-20$

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(C) Let the angle between the tangents be $2	heta = \cos^{-1}\left(\frac{7}{16}ight)$.Then $\cos(2	heta) = \frac{7}{16}$.Using the identity $\cos(2	heta) = 2\cos^2	heta - 1$,we get $2\cos^2	heta - 1 = \frac{7}{16}$,which implies $2\cos^2	heta = \frac{23}{16}$,so $\cos^2	heta = \frac{23}{32}$.Thus,$\sin^2	heta = 1 - \frac{23}{32} = \frac{9}{32}$,and $	an^2	heta = \frac{9/32}{23/32} = \frac{9}{23}$.For a circle $x^2 + y^2 + 4x + 4y + c = 0$,the center is $O(-2, -2)$ and radius $r = \sqrt{(-2)^2 + (-2)^2 - c} = \sqrt{8 - c}$.The distance from point $P(2, 2)$ to center $O(-2, -2)$ is $d = \sqrt{(2 - (-2))^2 + (2 - (-2))^2} = \sqrt{4^2 + 4^2} = \sqrt{32}$.In the right-angled triangle formed by the point,center,and point of tangency,$	an	heta = \frac{r}{\sqrt{d^2 - r^2}}$.So,$	an^2	heta = \frac{r^2}{d^2 - r^2} = \frac{8 - c}{32 - (8 - c)} = \frac{8 - c}{24 + c}$.Equating the two expressions for $	an^2	heta$: $\frac{8 - c}{24 + c} = \frac{9}{23}$.$23(8 - c) = 9(24 + c) \implies 184 - 23c = 216 + 9c$.$-32c = 32 \implies c = -1$.Since the problem states two such circles exist,there might be a misinterpretation of the geometry or parameters,but based on the provided equation,the sum of values of $c$ is $-1$.

The angle between the tangents drawn from the point $(2, 2)$ to the circle $x^2 + y^2 + 4x + 4y + c = 0$ is $\cos^{-1}\left(\frac{7}{16}\right)$. If two such circles exist,then the sum of the values of $c$ is:

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